# Behavior of the hypergeometric function near x=1

It is known that the hypergeometric function $${}_2F_1(a, b, c; x)$$ defined by the series $$\sum_{n=0}^\infty \frac{a(a+1)\cdots (a+n-1)\cdot b(b+1)\cdots (b+n-1)}{c(c+1)\cdots (c+n-1) n!}x^n$$ behaves like $$(1-x)^\alpha$$ near $$x=1$$. I am wondering if there is a formula for the $$\alpha$$ in terms of the constants $$a,b$$ and $$c$$. Or, if there exists a criterion determining if $$\alpha$$ is an integer.

It would be nice if a reference is provided.

if $$\alpha=-a-b+c\in(0,1)$$, one has the expansion to leading order $${}_2F_1(a, b, c; x)=\frac{\pi \Gamma (c) }{\Gamma (c-a) \Gamma (c-b) \Gamma (\alpha+1)\sin \pi \alpha}$$ $$\qquad-(1 - x)^\alpha\frac{\pi \Gamma (c) }{\Gamma (a) \Gamma (b) \Gamma (\alpha+1)\sin \pi \alpha}+{\cal O}(1-x).$$

Here is one reference and another reference.

• Thanks for the answer. Do you know what happens when $c-a-b$ is out of $(0, 1)$? 2 days ago
• if $\alpha=c-a-b>1$, the term of order $1-x$ is the leading order term; if $\alpha<0$ but not integer the leading order term $(1-x)^\alpha$ is a pole at $x=1$. 2 days ago
• I am a little bit confused. If $\alpha = c-a-b > 1$ do you mean that the term $(1-x)^\alpha$ to be the leading order term? If $\alpha< 0$ but not an integer, why is $(1-x)^\alpha$ gives a pole? Also can you provide a reference? 2 days ago
• I added another reference that has these cases explicitly. 2 days ago
• Thanks! The second reference suffices. 2 days ago

The command of Mathematica 14.0

FullSimplify[Series[Hypergeometric2F1[a, b, c, z], {z, 1, 1}] // Normal]

results in $$\pi \Gamma (c) \csc (\pi (a+b-c)) \left(\frac{\left(\frac{(z-1) (a-c) (b-c)}{a+b-c-1}+1\right) (1-z)^{-a-b+c}}{\Gamma (a) \Gamma (b) \Gamma (-a-b+c+1)}+\frac{a b (z-1)}{\Gamma (c-a) \Gamma (c-b) \Gamma (a+b-c+2)}-\frac{1}{\Gamma (c-a) \Gamma (c-b) \Gamma (a+b-c+1)}\right).$$

• How does this answer the question? 2 days ago
• @SamHopkins: The question is " if there is a formula for the α in terms of the constants a,b and c". MMA produces such a formula. Don't hesitate to ask for further explanation in need. 2 days ago